What is the integral of 1/(sqrt(x-1)*(x^2)) ?

The integral of 1/(sqrt(x-1)*(x^2)) is arctan(sqrt(x-1))+1/2 sin(2arctan(sqrt(x-1)))+C

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integral of 1/(sqrt(x-1)*(x^2))

\int \frac{1}{x - 1 \cdot ( x ^{2} )} d x

arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1)) + C

Solution steps

\int \frac{1}{x - 1 x ^{2}} d x

Apply u-substitution

: \int \frac{2}{( u ^{2} + 1 ) ^{2}} d u

= \int \frac{2}{( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= 2 \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= 2 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 2 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 2 \cdot \frac{1}{2} (arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1)))

Simplify

2 \cdot \frac{1}{2} (arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1))) : arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1))

= arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1))

Add a constant to the solution

= arctan (x - 1) + \frac{1}{2} sin (2 arctan (x - 1)) + C

\frac{d}{d x} (- x (2 x - y^{3}))

\int (4 + x) d x

e^{\frac{1}{θ}}

n = 0 \sum \infty \frac{3}{n n}

\int \frac{sin ( 2 x )}{1 + cos ^{2} ( 2 x )} d x

What is the integral of 1/(sqrt(x-1)*(x^2)) ?
The integral of 1/(sqrt(x-1)*(x^2)) is arctan(sqrt(x-1))+1/2 sin(2arctan(sqrt(x-1)))+C